ÌâÄ¿ÄÚÈÝ

ÒÑÖªÒ»¶¨Î¶ÈÏ£¬2X(g)£«Y(g)  ?mZ(g)¡¡¦¤H£½£­a kJ¡¤mol£­1(a>0)£¬ÏÖÓмס¢ÒÒÁ½ÈÝ»ýÏàµÈÇҹ̶¨µÄÃܱÕÈÝÆ÷£¬ÔÚ±£³Ö¸ÃζȺ㶨µÄÌõ¼þÏ£¬ÏòÃܱÕÈÝÆ÷¼×ÖÐͨÈë2 mol XºÍ1 mol Y£¬´ïµ½Æ½ºâ״̬ʱ£¬·Å³öÈÈÁ¿b kJ£»ÏòÃܱÕÈÝÆ÷ÒÒÖÐͨÈë1 mol XºÍ0.5 mol Y£¬´ïµ½Æ½ºâʱ£¬·Å³öÈÈÁ¿c kJ£¬ÇÒb>2c£¬Ôòa¡¢b¡¢mµÄÖµ»ò¹ØϵÕýÈ·µÄÊÇ(¡¡¡¡)
A£®m£½4B£®a£½b C£®a<D£®m¡Ü2
D
b>2c£¬ËµÃ÷¼×ÈÝÆ÷ÖÐXµÄת»¯ÂʶÔÓÚÒÒÈÝÆ÷ÖÐXµÄת»¯ÂÊ¡£ÒòΪ¼×ÈÝÆ÷ÖÐXºÍYµÄÎïÖʵÄÁ¿¾ùÊÇYÈÝÆ÷µÄ2±¶£¬Õâ˵Ã÷Ôö´óѹǿƽºâÊÇÏòÕý·´Ó¦·½ÏòÒƶ¯µÄ£¬¼´m£¼3¡£ÒòΪÊÇ¿ÉÄæ·´Ó¦£¬ËùÒÔÈë2 mol XºÍ1 mol Y·´Ó¦£¬´ïµ½Æ½ºâ״̬ʱ£¬·Å³öÈÈÁ¿²»¿ÉÄÜÊÇa kJ£¬¼´a£¾b¡£´ð°¸ÊÇD¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(10·Ö£©ÀûÓùâÄܺ͹â´ß»¯¼Á£¬¿É½«CO2ºÍH2O(g)ת»¯ÎªCH4ºÍO2¡£×ÏÍâ¹âÕÕÉäʱ£¬ÔÚ²»Í¬´ß»¯¼Á£¨I ,II£¬III)×÷ÓÃÏ£¬CH4µÄ²úÁ¿Ëæ¹âÕÕʱ¼äµÄ±ä»¯ÈçÏÂͼËùʾ¡£

(1) ÔÚO?30СʱÄÚ£¬CH4µÄƽ¾ùÉú³ÉËÙÂʺÍÓÉ´óµ½Ð¡µÄ˳ÐòΪ_________£»·´Ó¦¿ªÊ¼ºóµÄ15СʱÄÚ£¬ÔÚµÚ_________ÖÖ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬ÊÕ¼¯µÄCH4×î¶à¡£
(2) ½«ËùµÃCH4ÓëH2O(g)ͨÈë¾Û½¹Ì«ÑôÄÜ·´Ó¦Æ÷£¬·¢Éú·´Ó¦
CH4(g)+H2O(g)CO(g) +3H2(g) ¡÷H=+206kJ¡¤mol-1¡£½«µÈÎïÖʵÄÁ¿µÄCH4ºÍH2O(g)³äÈë1LºãÈÝÃܱÕÈÝÆ÷£¬Ä³Î¶ÈÏ·´Ó¦´ïµ½Æ½ºâ£¬´Ëʱ²âµÃCOµÄÎïÖʵÄÁ¿ÎªO.10 mol,CH4µÄƽºâת»¯ÂÊΪ91 %£¬Ôò´ËζÈϸ÷´Ó¦µÄƽºâ³£ÊýΪ_________ (¼ÆËã½á¹ûÈ¡ÕûÊý£©¡£
(3) ¸Ã·´Ó¦²úÉúµÄCOºÍH2¿ÉÓÃÀ´ºÏ³É¿ÉÔÙÉúÄÜÔ´¼×´¼£¬ÒÑÖªCO(g)¡¢CH3OH¢ÅµÄȼÉÕÈÈ·Ö±ðΪºÍ£¬ÔòCH3OH(l)²»ÍêȫȼÉÕÉú³ÉCO(g)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³ÌʽΪ_________¡£
(4)¹¤ÒµÉϳ£ÀûÓ÷´Ó¦CO(g)+2H2(g)  CH3OH (g), ¡÷H<0ºÏ³É¼×´¼£¬ÔÚ230¡ãC?270¡ãC×îΪÓÐÀû¡£ÎªÑо¿ºÏ³ÉÆø×îºÏÊʵÄÆðʼ×é³É±Èn(H2)£ºn(C0),·Ö±ðÔÚ230¡ãC¡¢2500CºÍ2700C½øÐÐʵÑ飬½á¹ûÈçͼ¡£

¢Ù2700CµÄʵÑé½á¹ûËù¶ÔÓ¦µÄÇúÏßÊÇ_________ (Ìî×Öĸ£©£»
¢Ú2300Cʱ£¬¹¤ÒµÉú²úÊÊÒËáŠÓõĺϳÉÆø×é³Én(H2):n(CO)µÄ±ÈÖµ·¶Î§ÊÇ_________ (Ìî×Öĸ£©¡£
A. 1 ?1.5   B. 2. 5?3  C. 3. 5?4. 5
(5) ijͬѧÒÔʯīΪµç¼«£¬ÒÔKOHÈÜҺΪµç½âÖÊÉè¼Æ¼×´¼È¼Áϵç³Ø£¬Æ为¼«µÄµç¼«·´Ó¦Ê½Îª_________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø